# How to determine the radius of convergence of $\sum\frac{1}{n^{\sqrt{n}}}x^n$ without using l'hopital's rule?

How to determine the radius of convergence of $\sum\frac{1}{n^{\sqrt{n}}}x^n$ without using l'hopital's rule?

I first tried root test, so $|a_n|^{\frac{1}{n}}=\frac{1}{n^{\frac{1}{\sqrt{n}}}}$ which I cannot see what the limit it has.

Then I tried ratio test, so $R=\lim|\frac{a_n}{a_{n+1}}|=\lim|\frac{{(n+1)}^{\sqrt{n+1}}}{n^{\sqrt{n}}}|$ in which I have no idea how to cancel term and determine the limit.

Does anyone have idea?

• Where would L'Hopital be used here? – user228113 Jun 4 '17 at 16:54
• What is $x_n$? You probably mean $x^n$. – Mark Viola Jun 4 '17 at 17:02

$$n^{1/\sqrt n}=\left(n^{1/2\sqrt n}\right)^2=\left[\left(\sqrt n\right)^{1/\sqrt n}\right]^2\xrightarrow[n\to\infty]{}\ldots$$